263
Internat. J. Math. & Math. Sci.
VOL. 20 NO. 2 (1997) 263-266
ON A CONJECTURE OF VUKMAN
QING DENG
Department of Mathematics
Southwest China Normal University
Chongqing 630715, P.R CHINA
(Received October 27, 1993 and in revised form October 30,1995)
ABSTRACT. Let R be a ring A bi-additive symmetric mapping d R R R is called a symmetric
hi-derivation if, for any fixed y E R, the mapping z
D(z, 1) is a derivation The purpose of this paper
is to prove the following conjecture of Vukman
Let R be a noncommutative prime ring with suitable characteristic restrictions, and let
D R x R R and f z D(z, z) be a symmetric hi-derivation and its trace, respectively Suppose
that f(x) E Z(R)for all x R, where .fk+l(X) [.fk(X),X] for k > 1 and f](x) f(x), thenD =0
KEY WORDS AND PHRASES: Prime ring, centralizing mapping, symmetric bi-derivation.
1991 AMS SUBJECT CLASSIFICATION CODES: Primary 16W25; Secondary 16N60
INTRODUCTION
Throughout this paper, R will denote an associative ring with center Z(R). We write [z,y] for
xy- /x, and I, for the inner derivation deduced by a A mapping D R R
R will be called
symmetric if D(x, ) holds for all pairs x, y R A symmetric mapping is called a symmetric biderivation, if D(z + z) D(z, z) + D(y, z) and D(xy, z) D(:r, z)u + xD(u, z) are fulfilled for all
R defined by f(x) D(z,x) is called the trace of the symmetric
z,/ R The mapping f :R
bi-derivation D, and obviously, f (z + 1) f (x) + f (y) + 2D(z, y) The concept of a symmetric biderivation was introduced by Gy Maksa in [1,2]
Some recent results cnceming symmetric
bi-derivations of prime tings can be found in Vukman [3,4]. In [4], Vukman proved that there are no
nonzero symmetric bi-derivations D in a noncommutative prime ring R of characteristic not two and
three, such that [[D(x, x), x], x] Z(R). The following conjecture was raised Let R be a
noncommutative prime ring of characteristic different from two and three, and let D R x R
R be a
symmetric bi-derivation. Suppose that for some integer n > 1, we have f, (x) E Z(R) for all z R,
1.
,
where fk+l(z) [fk(x),x] for k 1,2
and fl(z) D(x,x) ThenD 0.
The purpose of this paper is to prove this conjecture under suitable characteristic restrictions
2. TIlE RESULTS
THEOREM 1. Let R be a prime ring of characteristic different from two Suppose that R admits a
nonzero symmetric bi-derivation. Then R contains no zero divisors.
PROOF. It is sufficient to show that, a
0 for a R implies a 0 We need three steps to
establish this
LEMMA A. If D(a, .) g: 0, then D(a, .) #I,, where # C, the extended centroid of R
PROOF. Since D(a x) D(0, z) 0, we have
:,
264
QING DENG
aD(a,x)+D(a,x)a=O for all xR.
Replacing x by xy, we obtain
Ia(x)D(a,y)
D(a,x)Ia(y)
R;
for all z
and replacing y by y z, we get
Ia(z)yD(a,z)
D(a,x)yI(z),z,y,z
R.
(2.1)
Since D(a, .) -7= 0, we may suppose that D(a,z) 0 for a fixed z R. Obviously I(Z) :t/: 0 By
(2 1), and by [5, Lemma 1.3.2], there exist #(x) and t/(x) in C, either #(x) or t:(x) being not zero, such
that :z(x)Ia(x) + (x)D(a,x) 0. If (z)
(z) Ia(x); on the other hand, if
(x) 0 then # (x)I (x) 0 and I (x) 0, using (2.1) and Ia (z) 0, so D (a, x) 0. In any event,
we have D(a,z)= :z(x)I(x) Hence (2.1) implies (#(x)- #(z))I(x)yI(z)= 0 It follows that
either I(x) 0 or (x) #(z) By (2.1), the former implies D(a,z) 0 and D(a,x) #(z)I(x)
In both cases, we get D(a,x) #(z)I(x) for all x R, and 0 (z) being fixed
The fixed element # in Lemma A is somewhat dependem on a, we write it as # For any given
r Rara satisfies our original hypotheses on a; therefore for each r R, either D(ara, .)= 0 or
d(ara, .) Iaalaa, where aa O.
LEMMA B. If D(ara, ,) O, then #,,,,
PROOF. D(ara,,):O implies araO Suppose that D(a, ,) 0, then D(ara, z)=D(a,z)ra+
aD(r,x)a + arD(a,x) aD(r,x)a; but D(ara, x) ,a,,I,,a(x) =/,,a(arax xara), so that
:
lr=(arax xara) aD(r,x)a Right-multiplying the last equation by a, we have araaraxa 0 for
all x R. It follows that ara 0, a contradiction Therefore D(a, .) #,I,, and consequently,
D(. )
.oZ() + D(. ) + ,o();
and right-multiplying this equation by a yields
D(ara, x)a =/aaraxa for all
x
R.
Hence #aaaraxa #aaraxa, immediately #aa #a.
LEMMA C. If a
0, then a 0.
PROOF. Let S {r R lD(ara,.) =/rIa,,#,a 0} and T {r R \ D(ara,.) 0}
By Lemma A and B, R S 12 T and S and T are additive subgroups of R We conclude that either
S=RorT=R.
Suppose that S R Lemma A gives, either D(a, .) 0 or )(a, .) #I. If D(a, .) 0, then
D(ara, x) aD(r,x)a, for all r,x E R, and D(ara, x)a 0. It follows that :zaaraxa 0. Since
0, we have a 0 IfD(a, .) #aIa, then the equation
#a
#
:
D(. u)
D(. U). + D(.. u) + D(. U)
gives #araya 2/.tayara + taraya. Hence we get ayara 0, and a 0 again
We suppose henceforth that T R If D(a, .)= O, then D(axa, yz)= aD(xa, yz)= 0, and
ayD(xa, z) 0. Thus D(xa, z) D(x, z)a 0, and D(x, y)za D(x, yz)a 0 Since D :/- 0, we
then get a 0. If D(a, .) #aI, then, right-multiplying the equation D(axa, y) 0 by a, we obtain
axD(a,y)a 0, and a 0 again. The proof of the theorem is complete
/aaxaya
In order to prove Vukman’s conjecture, we need the following proposition.
PROPOSITION. Let n be a positive integer; let R be a prime ring with char R 0 or char R > n;
and let g be a derivation of R and f the trace of a symmetric hi-derivation D For
1, 2, n, let
F,(X,Y,Z) be a generalized polynomial such that, F(kx, f(kx),g(kx))= k’F,(x,f(z),g(x)) for all
x R for k 1, 2,
n. Let a R, and (a) the additive subgroup generated by a If for all x (a),
265
ON A ONJECTURE OF VUKMAN
F,,(z,f(z),9(z)) + Fr,-,(z,f(z),9(z)) +... + F(z,f(z),9(z))
Z(R),
(22)
thenF,(a,f(a),g(a))
Z(R) for/= 1,2,...,n
This proposition can be proved by replacing z by a, 24, na in (2.2) and applying a standard "Van
der Monde argument
TItEOREM 2. Let n be a fixed positive integer and R be a prime ring with char R 0 or char
R > n +2 Let fk+l(X)--[fk(x),x] for k > 1, and k(x)= f(x) the trace of a symmetric biderivation D ofR. Iff,(x) E Z(R) for all x E R, then either D 0 or R is commutative
PROOF. Linearizing f,.,(x) Z(R), we obtain
[[...[f(x) + f(y) + 2D(x,y),x y]
x
+ y],x + y] Z(R);
and using the Proposition, we get
[...[[f(x), y], x],
x]-+-[...[[f(x),x],y], ...x] + ...-+-[...[f(x),x],...y]
+ 2[...[[D(x,y),x],x],...,x] e Z(R),
equivalently,
(- 1)-2I-2([f,(x),y]) + (- 1)n-31x-3([fz(x),y]) -]-...
+ [fn-l(X),y] + 2(- 1)n-llz-l(D(x,y)) Z(R).
Noting that
(-
(23)
....
(-
[fn-l(X),X 2] --(-- 1)n-1/Tz-’ (D(x, xg))
and replacing y by x in (2.3), we then get 2(n +
1)f,(x)x
Z(R)
2f.(x)x,
z(R), it follows that
Since f(x)
A(=) =o
The lineafization of f (z)
0 gives
(- 1)-2I-l([f(x),y]) + (- 1)-aI2-a([A(x),vl)
+ + [A_l(X), V] + 2(- 1)n-lI-l(D(x,v))
- - -
I-([h-(x),xy]) xI-([_(x),y]) + I-k(h(x)y) for k 2,3
I (D(x, xy)) xI (D(x, y)) + I (f (x) y). Substituting xy for y in (2 4), we have
(-- 1)n-2I:-2(f2(x)y) + (-- 1)n-aIg-3(f3(x)y + + (-- 1)
(I=(f-x(X)y) + 2(- 1)-)I-l(fa(x)y) O.
Since
f-2(x), applfing I:(ab)=
Tang y
=0
(k)i:_a(a)i(b)d
J
noting
(2.4)
0.
I(f,(x))=O for
n,
+j
d
> n,
we then conclude that
2(_1)n_
ut (-
(n--1)_2(fl(Z)iz(fn_()))W(_l)n_ (n--2)_3(f2(z))iz(fn_,())+
1
)-’(A-(=))I=(A_=(z)) (A_I(=)) =,
1
+ (- )A_(z)z(A_=(z))
y
(n + 2)(n- 1)(A_(z)) 0,
0.
the
so
on the chactefistic, we get (f._l(Z)) =0 Suppose that D # 0 By Theorem 1,
f._(z)=O, and by induction, k(z)= [f(z),z] =0 Using Vukm [3, Theorem 1], R is
coutative, we complete the proof of Theorem 2
EOM 3. Let n > 1 be m imeger d R be a prime ring with ch R 0 or ch R > n + 1,
and let f(z) be the trace of a setfic bi-defivation D of R Suppose that [z f(z)] e Z(R) for 1
z R In ts case either D 0 or R is coutative
hotheses
a,
266
Q1NG DENG
PROOF. Using the condition Ix n, l(x)] E Z(R), we get [x 2n, f(x2)] E Z(R), and
Ix ,
Noting that
[’, s()] + [’, s()] + 2[, s)] z(n),
Ix f(z)] 2[z =, f(z)lz we now have om (2.5) that Six", f(z)lz +
,
,
Z(R) Thus
0 or z +
f(z)]
Z(R).
But lineafizing [z ", f(z)] Z (R) d applng the Proposition gives
[Xn-1 v + xn-2vx + + VX n-l, f(x)] + 2[x",D(x,v)] e Z(R)
either
x 3, elds
for all x, y e R, d tng y
,
-
,
Suppose that [x f(x)] # 0, then x +2 e Z(R) and Ix f(x)]x e Z(R), hence x e Z(R) Now this
0 or x
condition, together th x +2 Z(R), implies either x
Z(R), so that in each event,
[, y()] =0
Linemzing
Ix
,
f(x)]
0 and using the Proposition, we have
[-1 +
+ + v-x, y()] + 2[, D(, )]
0
Replacing y by x elds n[x+,f(x)] 0, hence [x,f(x)]x= 0 If D # 0, then by Theorem l,
Ix, f(x)] 0, d by Vukman [3, Theorem 1], R is commutative TMs completes the proof
ACOWLEDGMENT.
indebted to Prof M. N DMf for his help
referee for Ms vMuable suggestions.
would Mso like to th the
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,
,
303-307
J., Setfic bi-defivations on prime and sepfime tings, Aeationes Math. 3
(989), 245-254
J, Two results conceng setfic bi-defivations on prime tings, Aeatones Math.
[4]
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[5] RS, I.N., Rings wtth Invo&tton, Uversity of CMcago Press, 1976.
[3]